Work Energy Theorem: Practice Problem & Solution
A body of mass $1\text{ kg}$ is thrown upwards with a velocity $20\text{ m/s}$. It momentarily comes to rest after attaining a height of $18\text{ m}$. How much energy is lost due to air friction? ($g = 10\text{ m/s}^2$) (2009)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Work Energy Theorem. Understanding the underlying formula is key to arriving at the correct answer below:
Apply conservation of energy with non-conservative forces: initial energy $E_i = \frac{1}{2}mv^2 = 200\text{ J}$. Final potential energy at $18\text{ m}$ is $U_f = mgh = 180\text{ J}$. Energy lost against air friction is $\Delta E = E_i - U_f = 200 - 180 = 20\text{ J}$.
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